Given two numbers represented as strings, return multiplication of the numbers as a string.
Note: The numbers can be arbitrarily large and are non-negative.
大整数乘法,通过模拟手算过程来完成。
把结果存在整型数组里,先做乘加再考虑进位问题。
/*Inspired by Manual Multiply calculation process*/ class Solution { public: string multiply(string num1, string num2) { int len1 = num1.size(); int len2 = num2.size(); if(num1=="0"||num2=="0") return "0"; vector<int> res(len1+len2,0); for(int i=0; i<len1; ++i) for(int j=0; j<len2; ++j) res[i+j+1] += (num1[i]-'0')*(num2[j]-'0'); //multiply but not carry string s = ""; for(int k = len1+len2-1; k>=0; --k){ //vector int change to string if(k>0) res[k-1] += res[k]/10; res[k] = res[k]%10; char temp = '0'+res[k]; s = temp+s; } return s[0]=='0'?s.substr(1):s; } };